Category Springer Texts in Business and Economics

We start by constructing a confidence interval for any linear combination of в, say c’ в. We know that cCвOLS ~ N(с’в,&2с'(X’X)-1c) and it is a scalar. Hence,

Zobs = (c’Pols – cc e)/o(cc(X’X )-1 c)1/2 (7.26)

is a standardized N(0,1) random variable. Replacing a by s is equivalent to dividing zobs by the square root of a x2 random variable divided by its degrees of freedom. The latter random variable is (n — k)s2/a2 = RSS/a2 which was shown to be a хП-к. Problem 8 shows that zobs and RSS/a2 are independent. This means that

tobs = (c%ls — c в)/s(c'(X’ X )-1c)1/2 (7.27)

is a N(0,1) random variable divided by the square root of an independent хП-к/(n — k). This is a t-statistic with (n — k) degrees of freedom. Hence, a 100(1 — a)% confidence interval for c’в is